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Question
Find the points on curve y = x3 – 6x2 + x + 3 where the normal is parallel to the line x + y = 1729
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Solution
y = x3 – 6x2 + x + 3
Differentiating w.r.t. ‘x’
Slope of the tangent `("d"y)/("d"x)` = 3x2 – 12x + 1
Slope of the normal = `1/(3x^2 - 12x + 1)`
Given line is x + y = 1729
Slope of the line is – 1
Since the normal is parallel to the line, their slopes are equal.
`1/(3x^2 - 12x + 1)` = – 1
3x2 – 12x + 1 = 1
3x2 – 12x = 0
3x(x – 4) = 0
x = 0, 4
When x = 0, y = (0)3 – 6(0)2 + 0 + 3 = 3
When x = 4, y = (4)3 – 6(4)2 + 4 + 3
= 64 – 96 + 4 + 3
= – 25
∴ The points on the curve are (0, 3) and (4, – 25).
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